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(x^2-14x+32)=0
We get rid of parentheses
x^2-14x+32=0
a = 1; b = -14; c = +32;
Δ = b2-4ac
Δ = -142-4·1·32
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{17}}{2*1}=\frac{14-2\sqrt{17}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{17}}{2*1}=\frac{14+2\sqrt{17}}{2} $
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